I'm taking general relativity at the moment, and today in class the instructor gave us a definition of tangent vector as:
$v$ is a tangent vector based at $p\in M$ if $v_{p}$ is a linear combination of the directional derivative $\frac{\partial}{\partial x^\mu}$ $$ v=v^\mu \frac{\partial}{\partial x^\mu} \bigg|_{p} $$
He also mention that vectors and directional derivatives are in one-to-one correspondence.
I don't see why this definition is equivalent to the old Calculus definition of tangent vector: That given a curve $\gamma(\tau): (-\epsilon,\epsilon) \rightarrow M$ with parameter $\tau \in (-\epsilon,\epsilon)$ and $\gamma(0)=p$, the components of the tangent vector to the curve at $p$ is
$$
v^{\mu}=\frac{\partial{x^\mu}}{\partial \tau}\bigg|_{\tau=0}
$$
His new definition of tangent vector is a sum, while the old definition is a tuple. I have not yet understood manifolds, so can you explain this in a simple way with examples such that a student of Physics can understand?
For $v\in T_pM$, we want to view $v$ as a directional derivative operator. That is, if $c$ is a curve with $c'(0)=v$, then $$ v(f) : =\frac{d}{dt} f\circ c(t) $$
Here we must show that $v(f)$ is independent of curve $c$.
(1) Let $w:=(d\phi)^{-1}v $ Then if $c=\phi\circ \alpha,\ \alpha'(0)=w,$ $$ v(f)= \frac{d}{dt} f\circ c(t)=\frac{d}{dt} f\circ \phi\circ \alpha (t)=d(f\circ \phi)\ w $$ That is it is independent of curve. But we must consider $\phi,\ w$
(2) $\psi$ is another chart. Then $$ v(f)= d(f\circ \psi)\ x,\ x:=(d\psi)^{-1}(v) $$
Here $ d(\psi^{-1}\circ \phi)\ w=x $ so that $d(f\circ \psi)\ x=d(f\circ \phi)\ w $. That is it is independent of charts.